Class Assignment Question

#1

I am unable to put the condition for repeatition . Please help me with my code.
import java.util.*;
class Main{
public static Scanner scn = new Scanner(System.in);
public static void main(String [] args){

    // SUDOKU();
    // nqueen();
    // rat();
    // funky();
    // tricky();
    assign();
}

public static void assign(){
    int t = scn.nextInt();
    while(t-->0){
        int n = scn.nextInt();
        System.out.println(ca(n,"ab",""));
    }
}

public static int ca(int n_dig,String ques,String ans){
    if(ans.length()==n_dig){
        System.out.println(ans);
        return 1;
    }


    int count = 0;

    for(int i = 0; i< ques.length(); i++){
        char cc = ques.charAt(i);
        if(cc == 'b'){
        count += ca(n_dig,ques,ans + 'a');
        }else{
        count +=    ca(n_dig,ques,ans + 'a');
        count +=    ca(n_dig,ques,ans + 'b');
        }
    }
    return count;
}

}

#2

@ap8730390,

Are you sure you have tagged the correct question? Are you attempting Class Assignment question or some other question? Because the code is very different.

#3

Yeah, I am trying to attempt Class Assignment question using recursion

#4

@ap8730390,

Suggested approach:

If observed carefully , we can identify that this is a problem for fibonacci series. This is because at the nth place , there are two possibilities.

Possibility 1 : We can choose to place the current character as ‘a’. If so , then it doesn’t matter whether we placed ‘a’ or ‘b’ at the previous position. The total number of ways in this possibility would equal to f(n-1)

Possibility 2 : We can place the current character as ‘b’. However we can only do it if the previous character was not ‘b’ . Hence the total number of ways for this case must be f(n-2)

We add these two possibilities up and obtain the recursive relation
f(n) = f(n-1) + f(n-2)
This is clearly the recursive relation for Fibonacci Series .

For n=1 i.e. string of length 1
{a,b}=2
(First base case)

For n=2 i.e. string of length 2
{aa,ab,ba}=3 //b cannot be together
(second base case)

For n=3
{aaa,aba,baa,aab,bab}=5
fib[3]=fib[2]+fib[1]=3+2=5

For n=4
{aaaa,abaa,baaa,aaba,baba,aaab,abab,baab}=8
fib[4]=fib[3]+fiib[2]=5+3=8

so on…
Concluding: fib[n]=fib[n-1]+fib[n-1]

#5

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